Where is $\Bbb R$ in the von Neumann hierarchy and the constructible hierarchy? Definitions


*

*$(a,b) := \{\{a\},\{a,b\}\}$

*$\Bbb N := \{n \in \infty \mid \forall A (0 \in A \land \forall n (n \in A \to n \cup \{n\} \in A) \to n \in A) \}$, where $\infty$ is the set guaranteed to exist by the axiom of infinity.

*$\Bbb Z := \Bbb N \times \Bbb N / \sim$ where $(a,b) \sim (c,d) \iff a+d=b+c$

*$\Bbb Q := \Bbb Z \times (\Bbb N \setminus \{0\}) / \sim$ where $(a,b) \sim (c,d) \iff ad = bc$

*$\Bbb R := \{S \subseteq \Bbb Q \mid \forall x \forall y (x \in S \land y < x \to y \in S) \land \\ \exists m (m \in \Bbb Q \land \forall x (x \in S \to x < m)) \land \\ \forall m (m \in S \to \exists n (n \in S \land m < n)) \}$


So $\Bbb R$ is constructed using Dedekind cuts.
Examples
We have:
$$\varnothing = 0_\Bbb N \in \{0_\Bbb N\} \in (0_\Bbb N,0_\Bbb N) \in 0_\Bbb Z \in \{0_\Bbb Z\} \in (0_\Bbb Z,1_\Bbb N) \in 0_\Bbb Q \in 1_\Bbb R \in \Bbb R$$
Also:
$$\varnothing = 0_\Bbb N \in 1_\Bbb N \in \{1_\Bbb N\} \in (0_\Bbb Z,1_\Bbb N) \in 0_\Bbb Q \in 1_\Bbb R \in \Bbb R$$
This might faciliate the calculation.
Question
Where is $\Bbb R$ in the von Neumann hierarchy and the constructible hierarchy?
In other words, what is the least ordinal $\alpha$ and the least ordinal $\beta$ such that $\Bbb R \in L_\alpha$ and $\Bbb R \in V_\beta$?
 A: The question of where $\mathbb{R}$ lies in the cumulative hierarchy is sensitive to the exact definitions we use; it will always be in some $V_{\omega+n}$ for $n$ finite and small, but the precise value of $n$ may vary between reasonable choices of definition. At a glance I think your definition gives $n=6$ (Henning's calculation looks right to me), but I could be missing something.

However, the situation with regards to the constructible hierarchy is much more canonical and interesting:


*

*First of all, it is consistent that $\mathbb{R}\not\in L$: this will happen exactly when there is a non-constructible real.

*So really you want to ask about $L$'s version of the real numbers, $\mathbb{R}^L$. It turns out that this set shows up first in $L_{\omega_1^L+1}$ (where "$\omega_1^L$" is the least ordinal which $L$ thinks is uncountable - we may of course have $\omega_1^L<\omega_1$, just as we may have $\mathbb{R}^L\subsetneq\mathbb{R}$). The proof goes roughly:


*

*First, by the condensation lemma every real in $L$ is in $L_{\omega_1^L}$.

*Since being a real is a definable property, this means that $\mathbb{R}^L$ is a definable subset of $L_{\omega_1^L}$ - that is, $\mathbb{R}^L\in L_{\omega_1^L+1}$.

*Finally, it can be shown (see the discussion here) that for every $\alpha<\omega_1^L$ there is a real $r$ in $L$ but not in $L_\alpha$. This shows that $\mathbb{R}^L\not\in L_\alpha$ for any $\alpha<\omega_1^L$, and so $\mathbb{R}^L\not\in L_{\omega_1^L}$ either.

*Combining the previous two bullet points, we have that $\mathbb{R}^L$ appears in the $L$-hierarchy exactly at stage $\omega_1^L+1$.
